For a start, i would tend to agree that the objective probability is 1/3, 2/3 as mentioned in the above threads. However, my stand is that it does not make a difference whether he switches or not.
This is my explanation:
Let us look at the state space in actual: Given three boxes A, B, C.
There are two main events in general that can happen:
1) Player chooses the correct box prior -
Say the keys are in A and player (P) chooses A. Host (H) chooses B. The actions are illustrated as follows:
P = A, H = B ---- switch => lose
Similarly, P = A, H = B ---- switch => lose
If you account for all permutations for the above event of (1), the keys can be in box A, B or C and player chooses the right box. Hence, for (1), and six events in total in which the player loses if he switch.
2) Player chooses the box without the keys prior
Say the keys are in A and player chooses B. Host can only choose C. The actions are illustrated as follows:
P = B, H = C ---- switch => win
P = C, H = B ---- switch => win
For the three possible positions of the keys, six events in total in which the player wins if he switches .
The state space has 12 events in all possibility. Six in which he will lose if he switches, six in which he wins if he switches. Hence no nash equilibrium. There is no action that is superior to the other.
Given also the lack of nash equilibrium on the participants perspective, no baynesian equilibrium on the part of the host. He cannot exploit his informational superiority and lead the participant into losing.
Just my half cents worth... Dunno right or not